$limsup_{ntoinfty}{a_n}^{b_n}= limsup_{ntoinfty}{a_n}^{limlimits_{ntoinfty}b_n}$ under certian conditions.












0












$begingroup$


I have a proof that requires the following justification. So, I decided to recast it in the following form:




If $limlimits_{ntoinfty}b_n$ exists and $limsup_{ntoinfty}{a_n}=a>0$, then $limsup_{ntoinfty}{a_n}^{b_n}= limsup_{ntoinfty}{a_n}^{limlimits_{ntoinfty}b_n}$




My proof



begin{align} limsup_{ntoinfty}{a_n}^{b_n}&=limsup_{ntoinfty}e^{ln{a_n}^{b_n}}\&=limsup_{ntoinfty}e^{b_nln{a_n}}\&=e^{limsup_{ntoinfty}(b_nln{a_n})}\&=e^{limlimits_{ntoinfty}b_nlimsup_{ntoinfty}ln{a_n}}\&=e^{limlimits_{ntoinfty}b_nln{(limsup_{ntoinfty}a_n)}}\&=e^{limlimits_{ntoinfty}b_nln{a}}\&=e^{ln{a}^{limlimits_{ntoinfty}b_n}}\&={a}^{limlimits_{ntoinfty}b_n}\&={limsup_{ntoinfty}{a_n}}^{limlimits_{ntoinfty}b_n}end{align}
Please, is this correct? And is there any other proof?






real-analysis limsup-and-liminf







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  • $begingroup$
    Proofs shouldn't just be equations, they should be sentences. For example, a sentence about why line $2$ and line $3$ are equal...
    $endgroup$
    – 5xum
    Jan 30 at 9:39










  • $begingroup$
    @5xum: Sorry, it was a typo.
    $endgroup$
    – Micheal
    Jan 30 at 9:45










  • $begingroup$
    @5xum: I edited it.
    $endgroup$
    – Micheal
    Jan 30 at 11:37
















0












$begingroup$


I have a proof that requires the following justification. So, I decided to recast it in the following form:




If $limlimits_{ntoinfty}b_n$ exists and $limsup_{ntoinfty}{a_n}=a>0$, then $limsup_{ntoinfty}{a_n}^{b_n}= limsup_{ntoinfty}{a_n}^{limlimits_{ntoinfty}b_n}$




My proof



begin{align} limsup_{ntoinfty}{a_n}^{b_n}&=limsup_{ntoinfty}e^{ln{a_n}^{b_n}}\&=limsup_{ntoinfty}e^{b_nln{a_n}}\&=e^{limsup_{ntoinfty}(b_nln{a_n})}\&=e^{limlimits_{ntoinfty}b_nlimsup_{ntoinfty}ln{a_n}}\&=e^{limlimits_{ntoinfty}b_nln{(limsup_{ntoinfty}a_n)}}\&=e^{limlimits_{ntoinfty}b_nln{a}}\&=e^{ln{a}^{limlimits_{ntoinfty}b_n}}\&={a}^{limlimits_{ntoinfty}b_n}\&={limsup_{ntoinfty}{a_n}}^{limlimits_{ntoinfty}b_n}end{align}
Please, is this correct? And is there any other proof?






real-analysis limsup-and-liminf







share|cite|improve this question











$endgroup$












  • $begingroup$
    Proofs shouldn't just be equations, they should be sentences. For example, a sentence about why line $2$ and line $3$ are equal...
    $endgroup$
    – 5xum
    Jan 30 at 9:39










  • $begingroup$
    @5xum: Sorry, it was a typo.
    $endgroup$
    – Micheal
    Jan 30 at 9:45










  • $begingroup$
    @5xum: I edited it.
    $endgroup$
    – Micheal
    Jan 30 at 11:37














0












0








0


1



$begingroup$


I have a proof that requires the following justification. So, I decided to recast it in the following form:




If $limlimits_{ntoinfty}b_n$ exists and $limsup_{ntoinfty}{a_n}=a>0$, then $limsup_{ntoinfty}{a_n}^{b_n}= limsup_{ntoinfty}{a_n}^{limlimits_{ntoinfty}b_n}$




My proof



begin{align} limsup_{ntoinfty}{a_n}^{b_n}&=limsup_{ntoinfty}e^{ln{a_n}^{b_n}}\&=limsup_{ntoinfty}e^{b_nln{a_n}}\&=e^{limsup_{ntoinfty}(b_nln{a_n})}\&=e^{limlimits_{ntoinfty}b_nlimsup_{ntoinfty}ln{a_n}}\&=e^{limlimits_{ntoinfty}b_nln{(limsup_{ntoinfty}a_n)}}\&=e^{limlimits_{ntoinfty}b_nln{a}}\&=e^{ln{a}^{limlimits_{ntoinfty}b_n}}\&={a}^{limlimits_{ntoinfty}b_n}\&={limsup_{ntoinfty}{a_n}}^{limlimits_{ntoinfty}b_n}end{align}
Please, is this correct? And is there any other proof?






real-analysis limsup-and-liminf







share|cite|improve this question











$endgroup$




I have a proof that requires the following justification. So, I decided to recast it in the following form:




If $limlimits_{ntoinfty}b_n$ exists and $limsup_{ntoinfty}{a_n}=a>0$, then $limsup_{ntoinfty}{a_n}^{b_n}= limsup_{ntoinfty}{a_n}^{limlimits_{ntoinfty}b_n}$




My proof



begin{align} limsup_{ntoinfty}{a_n}^{b_n}&=limsup_{ntoinfty}e^{ln{a_n}^{b_n}}\&=limsup_{ntoinfty}e^{b_nln{a_n}}\&=e^{limsup_{ntoinfty}(b_nln{a_n})}\&=e^{limlimits_{ntoinfty}b_nlimsup_{ntoinfty}ln{a_n}}\&=e^{limlimits_{ntoinfty}b_nln{(limsup_{ntoinfty}a_n)}}\&=e^{limlimits_{ntoinfty}b_nln{a}}\&=e^{ln{a}^{limlimits_{ntoinfty}b_n}}\&={a}^{limlimits_{ntoinfty}b_n}\&={limsup_{ntoinfty}{a_n}}^{limlimits_{ntoinfty}b_n}end{align}
Please, is this correct? And is there any other proof?






real-analysis limsup-and-liminf




real-analysis limsup-and-liminf






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share|cite|improve this question








edited Jan 30 at 9:51







Micheal

















asked Jan 30 at 9:38









MichealMicheal

26511




26511












  • $begingroup$
    Proofs shouldn't just be equations, they should be sentences. For example, a sentence about why line $2$ and line $3$ are equal...
    $endgroup$
    – 5xum
    Jan 30 at 9:39










  • $begingroup$
    @5xum: Sorry, it was a typo.
    $endgroup$
    – Micheal
    Jan 30 at 9:45










  • $begingroup$
    @5xum: I edited it.
    $endgroup$
    – Micheal
    Jan 30 at 11:37


















  • $begingroup$
    Proofs shouldn't just be equations, they should be sentences. For example, a sentence about why line $2$ and line $3$ are equal...
    $endgroup$
    – 5xum
    Jan 30 at 9:39










  • $begingroup$
    @5xum: Sorry, it was a typo.
    $endgroup$
    – Micheal
    Jan 30 at 9:45










  • $begingroup$
    @5xum: I edited it.
    $endgroup$
    – Micheal
    Jan 30 at 11:37
















$begingroup$
Proofs shouldn't just be equations, they should be sentences. For example, a sentence about why line $2$ and line $3$ are equal...
$endgroup$
– 5xum
Jan 30 at 9:39




$begingroup$
Proofs shouldn't just be equations, they should be sentences. For example, a sentence about why line $2$ and line $3$ are equal...
$endgroup$
– 5xum
Jan 30 at 9:39












$begingroup$
@5xum: Sorry, it was a typo.
$endgroup$
– Micheal
Jan 30 at 9:45




$begingroup$
@5xum: Sorry, it was a typo.
$endgroup$
– Micheal
Jan 30 at 9:45












$begingroup$
@5xum: I edited it.
$endgroup$
– Micheal
Jan 30 at 11:37




$begingroup$
@5xum: I edited it.
$endgroup$
– Micheal
Jan 30 at 11:37










2 Answers
2






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It's correct because $f(x)=e^x$ and $ln$ are continuous functions.



For all continuous in $a$ function $f$ if $limlimits_{nrightarrow+infty}a_n=a$ then:
$$lim_{nrightarrow+infty}f(a_n)=fleft(lim_{nrightarrow+infty}a_nright)=f(a).$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Your proof seems a bit more complicated than it should be. Since $a^b$ is continuous for all $a,bin mathbb{R}$, we have that $limsup_{nrightarrowinfty}a_n^{b_n} = limsup_{nrightarrowinfty}a_n^{limsup_{nrightarrowinfty} b_n} $. However, since $lim_{nrightarrowinfty} b_n$ exists, we have $limsup_{nrightarrowinfty}b_n=lim_{nrightarrowinfty}b_n$, from which your claim follows.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
      2






      active

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      active

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      1












      $begingroup$

      It's correct because $f(x)=e^x$ and $ln$ are continuous functions.



      For all continuous in $a$ function $f$ if $limlimits_{nrightarrow+infty}a_n=a$ then:
      $$lim_{nrightarrow+infty}f(a_n)=fleft(lim_{nrightarrow+infty}a_nright)=f(a).$$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        It's correct because $f(x)=e^x$ and $ln$ are continuous functions.



        For all continuous in $a$ function $f$ if $limlimits_{nrightarrow+infty}a_n=a$ then:
        $$lim_{nrightarrow+infty}f(a_n)=fleft(lim_{nrightarrow+infty}a_nright)=f(a).$$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          It's correct because $f(x)=e^x$ and $ln$ are continuous functions.



          For all continuous in $a$ function $f$ if $limlimits_{nrightarrow+infty}a_n=a$ then:
          $$lim_{nrightarrow+infty}f(a_n)=fleft(lim_{nrightarrow+infty}a_nright)=f(a).$$






          share|cite|improve this answer











          $endgroup$



          It's correct because $f(x)=e^x$ and $ln$ are continuous functions.



          For all continuous in $a$ function $f$ if $limlimits_{nrightarrow+infty}a_n=a$ then:
          $$lim_{nrightarrow+infty}f(a_n)=fleft(lim_{nrightarrow+infty}a_nright)=f(a).$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 30 at 9:57

























          answered Jan 30 at 9:49









          Michael RozenbergMichael Rozenberg

          109k1896201




          109k1896201























              1












              $begingroup$

              Your proof seems a bit more complicated than it should be. Since $a^b$ is continuous for all $a,bin mathbb{R}$, we have that $limsup_{nrightarrowinfty}a_n^{b_n} = limsup_{nrightarrowinfty}a_n^{limsup_{nrightarrowinfty} b_n} $. However, since $lim_{nrightarrowinfty} b_n$ exists, we have $limsup_{nrightarrowinfty}b_n=lim_{nrightarrowinfty}b_n$, from which your claim follows.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Your proof seems a bit more complicated than it should be. Since $a^b$ is continuous for all $a,bin mathbb{R}$, we have that $limsup_{nrightarrowinfty}a_n^{b_n} = limsup_{nrightarrowinfty}a_n^{limsup_{nrightarrowinfty} b_n} $. However, since $lim_{nrightarrowinfty} b_n$ exists, we have $limsup_{nrightarrowinfty}b_n=lim_{nrightarrowinfty}b_n$, from which your claim follows.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Your proof seems a bit more complicated than it should be. Since $a^b$ is continuous for all $a,bin mathbb{R}$, we have that $limsup_{nrightarrowinfty}a_n^{b_n} = limsup_{nrightarrowinfty}a_n^{limsup_{nrightarrowinfty} b_n} $. However, since $lim_{nrightarrowinfty} b_n$ exists, we have $limsup_{nrightarrowinfty}b_n=lim_{nrightarrowinfty}b_n$, from which your claim follows.






                  share|cite|improve this answer











                  $endgroup$



                  Your proof seems a bit more complicated than it should be. Since $a^b$ is continuous for all $a,bin mathbb{R}$, we have that $limsup_{nrightarrowinfty}a_n^{b_n} = limsup_{nrightarrowinfty}a_n^{limsup_{nrightarrowinfty} b_n} $. However, since $lim_{nrightarrowinfty} b_n$ exists, we have $limsup_{nrightarrowinfty}b_n=lim_{nrightarrowinfty}b_n$, from which your claim follows.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 30 at 9:57









                  Mark

                  10.4k1622




                  10.4k1622










                  answered Jan 30 at 9:55









                  UnexpectedExpectationUnexpectedExpectation

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